Using Borel Cantelli lemma to show that the set of convergence of non degenerate independent random variables has measure zero.

Question

I'm trying the following:

If $X_1,\dots, X_n,\dots, $ are non degenerate independent and identically distributed random variables; then

\begin{equation*} \mathbb{P}\left(X_n \text{ converge }\right)=0 \end{equation*} I tried a lot this exercise; but there is something that I'm not using. Firstly, I wrote the set of convergence as:

\begin{equation} \bigcap_{k=1}^{\infty}\bigcup_{N=1}^{\infty}\bigcap_{m=n+1}^{\infty}\lbrace \mid X_n-X_m\mid\le \frac{1}{k}\rbrace \end{equation}

At this point, I can't see how to use Borel Cantelli lemma, even I can't see where I have to use the hypothesis of independence. I tried to write the set ${\mid X_n-X_m\mid}$ and making a change in the index to obtain independent set, but it's not working.

By the Kolmogorov One Zero law one can show that the set of convergence of $X_n$ has probability either 0 or 1. Using this, I tried to obtain a contradiction if I suppose that $\mathbb{P}\left(X_n \text{ converge }\right)=1$.

The thing that I can't see how to use is the hypothesis that the variables are non degenerate.

I hope that you can provide a hint or something!

Thanks

EDIT: The set where the sequence $X_n$ converge is the same where $\limsup X_n$ and $\liminf X_n$ are equal. Since $\limsup X_n$ is measurable with respect to tail sigma-algebra, it has to be constant a.s. Thus, if $X_n$ converge a.s., $X_n$ is constant eventually a.s.; which is a contradiction. Is this "intuitive" argument valid?

Answer 1

Let $\alpha = P(X_1 = X_2)$.

If $\alpha < 1$, by the dominated convergence theorem, there exists $\epsilon>0$ such that for any $m \ne n$, $P(|X_m - X_n| > \epsilon) = P(|X_1 - X_2| > \epsilon) > 0$. So $\sum_{n} P(|X_{2n} - X_{2n+1}| > \epsilon) = \infty$. Now use the Borel-Cantelli Lemma.

If $\alpha = 1$, suppose for a contradiction that $X_1$ is not degenerate. Then there exists $c$ such that $P(X_1 > c)$ and $P(X_1 \le c)$ are both non-zero. But then $P(X_1 \ne X_2) \ge 2 P(X_1 > c) P(X_1 \le c) > 0$.